Problem: Which of the following numbers is a multiple of 6? ${62,80,97,98,102}$
The multiples of $6$ are $6$ $12$ $18$ $24$ ..... In general, any number that leaves no remainder when divided by $6$ is considered a multiple of $6$ We can start by dividing each of our answer choices by $6$ $62 \div 6 = 10\text{ R }2$ $80 \div 6 = 13\text{ R }2$ $97 \div 6 = 16\text{ R }1$ $98 \div 6 = 16\text{ R }2$ $102 \div 6 = 17$ The only answer choice that leaves no remainder after the division is $102$ $ 17$ $6$ $102$ We can check our answer by looking at the prime factorization of both numbers. Notice that the prime factors of $6$ are contained within the prime factors of $102$ $102 = 2\times3\times17 6 = 2\times3$ Therefore the only multiple of $6$ out of our choices is $102$. We can say that $102$ is divisible by $6$.